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Can you help me solve my mathematical problem?

  1. Jun 13, 2012 #1
    I'm definitely messing up somewhere :P

    The Larmor Equation is

    [tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)[/tex]

    I derived an equation stated as

    [tex]\nabla \times \vec{F}_{ij} \cdot \hat{n} = \nabla \times \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]

    How is put in a simpler way from the original derivation, is imagine we have what we began with

    [tex]\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}[/tex]

    Take the dot product of the unit vector on both sides gives

    [tex]\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]

    Now taking the curl of F is

    [tex]\nabla \times \vec{F}_{ij} \cdot \hat{n} = \nabla \times \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]

    In light of this equation however

    [tex]\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]

    I could rewrite the Larmor energy as

    [tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e}(\vec{F}_{ij} \cdot \hat{n}) L \cdot S[/tex]

    But I am sure many agree that is not very interesting.

    (and this is really starting to make my brain cells burst), taking the curl of a force gives a F/length, however, the unit vector in this equation

    [tex]\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]

    cancels these out and what I have again is the force again... yes? ... very circular... since this would be true, then we know what the force is anyway given earlier:

    [tex]\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n} [/tex]

    Now, just a moment ago I found the Larmor energy written as

    [tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{1}{r}\frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)[/tex]

    Notice the 1/r term which is not in my original case. If that where true, and we plug in my force example again

    [tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{1}{r}\frac{\partial V(r_{ij})}{\partial r_{ij}}\hat{n} ( L \cdot S)[/tex]

    (If I am doing this right) the unit vector would cancel out with the radius term and what would be left with is

    [tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)[/tex]

    Now if that wasn't confusing I don't know what is.

    I've either had it right from the beginning, or I have made a tiny mistake which is making a huge impact on my understanding of my own equation. Any help would be gladly appreciated.
     
  2. jcsd
  3. Jun 13, 2012 #2
    What a mess my brain got into. Assuming this is correct mind you.

    I found a solution to my problem (I think). The Larmor equation is

    [tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)[/tex]

    What I kept deriving was:

    [tex]\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]

    What I really needed was the original derivation

    [tex]\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}[/tex]

    Then taking the curl of F gives

    [tex]\nabla \times \vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}}[/tex]

    Which removes the unit vector because nabla again is 1/length.

    So what I think I have ended up with was the right derivation for the modified Larmor energy except for a dot product made on the unit vector after all, it would be

    [tex]\Delta H = \frac{2\mu}{\hbar Mc^2 e} (\nabla \times \vec{F}_{ij}) L \cdot S[/tex]


    What a mess my brain got into. Assuming this is correct mind you.
     
  4. Jun 13, 2012 #3
    Now, does this seem right?
     
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